Let the acute angle bisector of the two plane | vedclass

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(B) The equations of the planes are $P_{1}: x-2y-2z+1=0$ and $P_{2}: 2x-3y-6z+1=0$. The equation of the angle bisector is given by $\left|\frac{x-2y-2

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$\left(-2, 0, -\frac{1}{2}\right)$

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(B) The equations of the planes are $P_{1}: x-2y-2z+1=0$ and $P_{2}: 2x-3y-6z+1=0$. The equation of the angle bisector is given by $\left|\frac{x-2y-2z+1}{\sqrt{1^2+(-2)^2+(-2)^2}}\right| = \left|\frac{2x-3y-6z+1}{\sqrt{2^2+(-3)^2+(-6)^2}}\right|$. This simplifies to $\frac{x-2y-2z+1}{3} = \pm \frac{2x-3y-6z+1}{7}$. To determine the acute angle bisector,we check the sign of $a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (1)(2) + (-2)(-3) + (-2)(-6) = 2 + 6 + 12 = 20$. Since $20 > 0$,the negative sign gives the acute angle bisector. Thus,$\frac{x-2y-2z+1}{3} = -\frac{2x-3y-6z+1}{7}$. $7(x-2y-2z+1) = -3(2x-3y-6z+1)$. $7x-14y-14z+7 = -6x+9y+18z-3$. $13x-23y-32z+10 = 0$. Testing the point $\left(-2, 0, -\frac{1}{2}\right)$: $13(-2) - 23(0) - 32(-\frac{1}{2}) + 10 = -26 - 0 + 16 + 10 = 0$. Therefore,the point $\left(-2, 0, -\frac{1}{2}\right)$ lies on the plane $P$.

Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane $P$. Then which of the following points lies on $P$?

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